Perhaps we have heard two poker players discussing “combinations” want to get a better idea of what they are referencing.
In this guide, we’ll discuss combinations in poker and how they can be used to improve our strategic understanding of the game.
- What Is a Combination?
- Hand Combinations in Poker
- Calculating Combos in Hold’em
- Practice Questions with Combos
- Combinations in Mathematics
What Are Combinations?
There are two slightly different topics with the same name.
The definitions are as follows:
1. A singular instance of a certain poker hand (usually referring to the hole cards).
2. Mathematics. A selection of items from a collection where order doesn’t matter.
We’ll talk more about that mathematical definition a little later.
For now, let’s focus on the first definition, which is the most commonly used in the context of a discussion on poker.
Hand Combinations in Poker
The term “combination” (or “combo” for short) refers to the different ways in which we can make a specifictype of poker hand.
Example: We are dealt AK preflop in Hold’em. How many different combinations of AK are there?
If we were so inclined, we could list every possible way of making AK. But most players simply remember that there are 16 combinations of every unpaired hand - 12 are off-suit, and 4 are suited.
Example: We are dealt 66 preflop in Hold’em. How many different combinations of 66 are there?
Once again, we could list all the possible combinations of 66, but it’s far easier to simply remember that there are 6 combinations of every pocket pair in Hold’em.
- Unpaired hand - 16 preflop combos
- Unpaired off-suit hand - 12 preflop combos
- Unpaired suited hand - 4 preflop combos
- Pocket pair - 6 preflop combos
Calculating Combos in Hold’em
It’s easy enough to remember preflop combos, but things get a little more complicated when we need to take in to account card removal effect. (Various board cards are known, reducing the available combinations of different types of starting hand).
It’s therefore helpful if we have a system for calculating combos.
Let’s start with unpaired hands.
Combos of Unpaired Hands:
Card 1 : Number of available cards * Card 2: Number of available cards
Returning briefly to our AK preflop we can now see how our value of 16 was calculated.
In a standard deck, there are 4 available Aces and 4 available Kings.
Therefore: 4 * 4 = 16 total combos of AK
Let’s see a slightly tougher question where we need to take into account card removal effects.
Example: In Hold’em, the flop texture is Ac8d7h. How many combinations of AK are there?
Therefore: 3 * 4 = 12 total combos of AK.
Ok, now let’s learn the rule for pocket pairs.
Combos of Unpaired Hands:
Number of Available Cards * (Number of Available Cards -1)
Returning to our preflop question with the pocket 6s, there are four available Sixes in the deck.
Therefore: (4 * 3) / 2 = 6 combinations of pocket Sixes.
Let’s again see a slightly tougher question where we need to account for card removal effects.
Example: In Hold’em, the flop texture is Ac6d7h. How many combinations of 66 are there?
Since one of the Sixes is already out there, there are now only 3 Sixes remaining in the deck.
Therefore: (3 * 2) / 2 = 3 combinations of pocket Sixes.
Practice Questions with Combos
We should now be equipped to answer slightly more advanced questions regarding combinations. We’ll need basic arithmetic skills and some common sense.
Example: We hold pocket Aces preflop. How many combinations of QQ+/AK does our opponent have?
If we ignore the card removal effect temporarily, we know our opponent would have the following:
QQ+ = 18 combos (6 * 3)
AK = 16 combos
- for a total of 34 combos.
However, our opponent holds two of the Aces meaning that the number of combos will be affected due to card removal effects. This scenario is sometimes referred to as the “blocker effect”. The QQ and KK combos will remain at 6 combos each, but the AA and AK combos will be affected.
There is 1 combo of Aces left since we hold two of the Aces already.
(2 * 1) / 2 = 1 combo of Aces
There are 8 combos of AK (since there are 2 Aces and 4 Kings available)
4 * 2 = 8 combos of AK
So, if we list all of the combos we get the following:
AA – 1 combo
KK – 6 combos
QQ – 6 Combos
AK – 8 Combos
for a total of 21 combos.
That’s a significant difference in available combos due to card removal or “blocker” effects.
Example: In Hold’em, the flop texture is Ac8d7h. How many different ways of making a set are there?
It’s generally good to remember that there are always 3 combinations of each set available. That’s a total of 9 combinations of sets.
Let’s try something slightly more challenging.
Example: In Hold’em, the flop texture is Ac8d7h. How many different ways of making top pair are there?
This situation is where logic and common sense come in. We know that there are 12 combos of any individual top pair type hand such as AJ (4 * 3 = 12).
But how many different Ax hands are there that make top pair?
There are 13 card ranks in a deck, so if we exclude AA, A8 and A9, we must be left with 10 different types of Ax hand that make top pair, each with 12 combinations. That’s a total of 120 (10 * 12) different combinations of top pair on this texture.
That’s still a relatively comfortable calculation. When things get significantly more complex, it’s good to remember there are commercially available poker equity calculators that can show us precise numbers of combinations in complex scenarios.
Combinations in Mathematics
Now let’s visit our second definition of the term combinations.
Mathematics:A selection of items from a collection where order doesn’t matter.
So far, we have seen ways of calculating hole card combinations. This second definition will help us calculate the probability of various board runouts.
Let’s rewrite our definition to make it poker related.
Poker: A selection of cards from the deck where order doesn’t matter.
Let’s start with an example question.
It’s often said that there are 19,600 different possible flops in Hold’em. Demonstrate this mathematically using combinations.
Let’s start by looking at the mathematical formula for combinations.
Let’s define the values in the formula.
n = total number to choose from (i.e. the number of cards in the deck)
r = total number we are choosing (i.e. how many cards we draw from the deck)
C = stands for combinations
! = mathematical function known as “factorial”. (e.g. 5! Is 5 * 4 *3 *2 *1)
Let’s assign values to n and r.
n = 52 because there are 52 cards in the deck.
r = 3 because we are drawing 3 cards to generate a flop.
Now plug our numbers into the formula -
(49!) * 3!
It’s possible to plug these numbers directly into a calculator (using appropriate brackets), but it’s also viable to simplify the formula manually.
Although the simplification method is slightly outside the scope of this article, it’s not overly complicated, and we can simplify the above formula to the following -
52 * 51 * 50
The result is 22,100 possible flop combinations. Interesting. So why is it that a quick online search reveals 19,600 possible flop combinations? Could it be that those calculations are accounting for the fact that 2 cards in the deck are often known (i.e. our holecards)?
Let’s rerun the calculation.
n = 50 because 50 cards are remaining in the deck after our hole cards are dealt.
r = 3 because we are drawing 3 cards to generate a flop.
This time we get -
47! * 3!
Which simplifies to:
50 * 49 * 48
- which equals 19,600!
So, when players say there are 19,600 flops, they mean assuming 2 cards from the deck are known. We’ve learned that there are 22,100 different flops that an outside observer would experience (given that he doesn’t know any of the hole cards).
Let’s use our knowledge of combinations to see if we can answer a slightly more advanced question.
Example: We are dealt ThJh in Hold’em. What are our chances of flopping a flush?
We already know how many different flops there are (19,600) so our next goal should be to establish how many ways there are of making a heart flush assuming there are 11 hearts left in the deck.
How can we establish the number of different 3 card combinations that can be dealt from a selection of 11 cards?
Once again, this calculation is precisely what mathematical combinations are designed to do.
n = 11 because this is how many hearts there are left in the deck assuming we already hold 2hearts in our hand.
r = 3 because this is the number of cards used to generate a flop in Hold’em.
Let’s input the numbers into our formula -
8! * 3!
- which simplifies to
11 * 10* 9
= 165 ways of 3heart flops being dealt.
The probability of flopping 3 hearts is, therefore -
165 / 19,600 = 0.0084 or 0.84%
However, when we check with equity calculation software, it tells us that the chance of flopping a flush is 0.82% rather than 0.84%. Can you see what the discrepancy is?
Some of those three heart flops give us a straight flush. We need to know how many so that we can discount it from our total number of flush flops.
Flops that make the straight flush with ThJh -
That’s four different flops that we need to discount from our current total of 165.
161 / 19,600 = 0.0082 or 0.82%
There we go. We have arrived at 0.82%, the same value given to us by equity calculation software.
Let’s try one more.
Example: We are dealt AKo, what are our chances of flopping a straight?
Perhaps we are starting to see a pattern for the simplified version of the formula when using combinations to arrive at the likelihood of various flops.
The pattern is as follows -
Numbers of Card 1 * Numbers of Card 2 * Numbers of Card 3
In the case of attempting to make a straight, we very specifically need to hit a TJQ flop (although the order of the cards doesn’t matter).
There are 12 ways we can hit either a T, J or Q on the flop.
There are 8 ways we can hit either a J or Q assuming we hit a T on the flop.
(whichever flop card there will be 8 ways of catching part of the straight on the turn)
There are 4 ways we can catch the straight completing river card.
12 * 8 * 4
If we multiply the top half of the formula (12 * 8 * 4), we get 384. This solution is the number of different ways of flopping the straight if we assume the order of the cards is relevant (which is not).
Therefore, 384 would suggest that TcJhQd and JhTcQd are different flops,which they are not for the purposes of this calculation. The mathematical name for a selection where the order matters is permutation rather than combination and involves a slightly different formula.
The second part of the formula (divide by 3! Or 6) is used to account for duplicate flops (where just the ordering is different).
That leaves us with 64 different ways of making a straight where the order of flop cards doesn’t matter.
Therefore: 64 / 19600 = 0.003265 or 0.33%
Using combinations to answer this question is not strictly necessary; we can confirm the answer by using the basic probability formula for the probability of successive events.
Let’s see how it works -
Event 1 – First flop card is either a T, J or Q = 12/50 because 50 cards remain in the deck.
Event 2- Second flop card is one of the remaining 8 cards to complete a straight = 8/49 because 49 cards remain in the deck.
Event 3 – Third flop card is one of the remaining 4 cards to complete a straight = 4/48 because 48 cards remain in the deck.
In probability theory, we simply multiply the probability of each successive event to establish the total probability.
12/50 * 8/49 * 4/48 = 0.003265
Look familiar? That’s the same value we calculated earlier for flopping a straight given that we are dealt AK preflop.
Hand Combinations Summary
The most commonly used type of combinations is the hand combinations we discussed at the outset. We should generally assume that this is what a player wants to discuss if they bring up the topic of combinations.
In contrast, far fewer players know about mathematical combinations. The average player may not even have heard of them, let alone know how to calculate them. Having a firm grasp of such mathematical concepts can hence give us a significant edge over the playing field.
We’ve learned that there are 19,600 different flops given two known hole cards. We can use combinations to establish how often certain types of flop hit, then divide the result by 19,600 to learn the respective probability.